How to resolve the algorithm Smith numbers step by step in the Racket programming language

Published on 12 May 2024 09:40 PM
#Racket

How to resolve the algorithm Smith numbers step by step in the Racket programming language

Table of Contents

Problem Statement

Smith numbers are numbers such that the sum of the decimal digits of the integers that make up that number is the same as the sum of the decimal digits of its prime factors excluding 1. By definition, all primes are excluded as they (naturally) satisfy this condition! Smith numbers are also known as   joke   numbers.

Using the number 166 Find the prime factors of 166 which are: 2 x 83 Then, take those two prime factors and sum all their decimal digits: 2 + 8 + 3 which is 13 Then, take the decimal digits of 166 and add their decimal digits: 1 + 6 + 6 which is 13 Therefore, the number 166 is a Smith number.

Write a program to find all Smith numbers below 10000.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Smith numbers step by step in the Racket programming language

Source code in the racket programming language

#lang racket
(require math/number-theory)

(define (sum-of-digits n)
  (let inr ((n n) (s 0))
    (if (zero? n) s (let-values (([q r] (quotient/remainder n 10))) (inr q (+ s r))))))

(define (smith-number? n)
  (and (not (prime? n))
       (= (sum-of-digits n)
          (for/sum ((pe (in-list (factorize n))))
            (* (cadr pe) (sum-of-digits (car pe)))))))

(module+ test
  (require rackunit)
  (check-equal? (sum-of-digits 0) 0)
  (check-equal? (sum-of-digits 33) 6)
  (check-equal? (sum-of-digits 30) 3)

  (check-true (smith-number? 166)))

(module+ main
  (let loop ((ns (filter smith-number? (range 1 (add1 10000)))))
    (unless (null? ns)
      (let-values (([l r] (split-at ns (min (length ns) 15))))
        (displayln l)
        (loop r)))))

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